Polycarp likes numbers that are divisible by 3.

He has a huge number $s$

.

For example, if the original number is $s=3121$

.

Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.

What is the maximum number of numbers divisible by $3$

Output

Print the maximum number of numbers divisible by $3$

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3121

2

6

1

1000000000000000000000000000000000

33

201920181

4

.

In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$

.

In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$

, $9$, $201$ and $81$ are divisible by $3$.

题意：给一个仅包含数字的字符串，将字符串分割成多个片段（无前导0），求这些片段里最多有多少是3的倍数

思路：一个数是3的倍数，则各位的和能被3整除。

对于单独的一个数字，如果是3的倍数，则ans++

否则，考虑连续的两个数字，如果是，则ans++

如果第三个数本身是3的倍数，则ans++

如果第三个数不是3的倍数，则对于前两个数的和对3取余，结果为[1,1]或者[2,2]（如果为[1,2],[2,1]，则这两个数字能够被3整除）

对于第三个数对3取余，结果为0,1,2

0：第三个数本身能被3整除ans++

1：[1,1,1]是3的倍数取全部，[2,2,1]取后两个   ans++

2：[1,1,2]取后两个 [2,2,2]是3的倍数，取全部  ans++

所以 对于n=3 一定可以找到

代码：

#include<cstdio>
#include<cstring>
using namespace std;
const int N = 2e5+5;
char str[N];
int main(){
	scanf("%s",str);
	int r=0,sum=0,ans=0,n=0;
	for(int i=0;i<strlen(str);i++){
		r=(str[i]-'0')%3;
		sum+=r;
		n++;
		if(r==0||sum%3==0||n==3){
			ans++;
			n=sum=0;
		}
	}
	printf("%d\n",ans);
	return 0;
}